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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2019 · 11 January · Evening Shift · Physics · Thermal Properties of Matter

A thermometer graduated according to a linear scale reads a value...

A thermometer graduated according to a linear scale reads a value x0x_{0} when in contact with boiling water, and x0/3x_{0} / 3 when in contact with ice. What is the temperature of an object in ∘C{ }^{\circ} \mathrm{C}, if this thermometer in the contact with the object reads x0/2?x_{0} / 2 ?A thermometer graduated according to a linear scale reads a value x_ 0 when in contact with boiling water, and x_ 0 / 3

Options

  1. A25
  2. B60
  3. C40
  4. D35

Correct answer

A. 25

Step-by-step solution

Let required temperature =T∘C=\mathrm{T}^{\circ} \mathrm{C} A thermometer graduated according to a linear scale reads a value x_ 0 when in contact with boiling water, and x_ 0 / 3
⇒T∘C=x02−x03=x06\Rightarrow \mathrm{T}^{\circ} \mathrm{C}=\frac{\mathrm{x}_{0}}{2}-\frac{\mathrm{x}_{0}}{3}=\frac{\mathrm{x}_{0}}{6} &(x0−x03)=(100−0∘C)\&\left(\mathrm{x}_{0}-\frac{\mathrm{x}_{0}}{3}\right)=\left(100-0^{\circ} \mathrm{C}\right)\Rightarrow \frac{2 \mathrm{x}_{0}}{3}=100 \Rightarrow \mathrm{x}_{0}=\frac{300}{2} \Rightarrow \mathrm{T}^{\circ} \mathrm{C}=\frac{\mathrm{x}_{0}}{6}=\frac{150}{6}=25^{\circ} \mathrm{C}
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